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Proof step 11 from proof prefixn-eq-prefixn-app (previous step)

n1 < # x0 ∧ n0 = n1 ∨ # x0 ≤ n0 ∧ # x0 ≤ n1 ∧ x1 = [] ∨ # x0 ≤ n0 ∧ n1 = # x0 ∧ x1 ≠ [] ⊦ (n0 ≤ # x0 ⊃ firstn(n0, x0);x0) = (n1 ≤ # x0 + # x1 ⊃ firstn(n1, x0 + x1);x0 + x1)

Heuristic if-then-else split applied rule case distinction

And obtained the following new goals

Proof step 12:
n0 ≤ # x0, n1 < # x0 ∧ n0 = n1 ∨ # x0 ≤ n0 ∧ # x0 ≤ n1 ∧ x1 = [] ∨ # x0 ≤ n0 ∧ n1 = # x0 ∧ x1 ≠ [] ⊦ firstn(n0, x0) = (n1 ≤ # x0 + # x1 ⊃ firstn(n1, x0 + x1);x0 + x1)
Proof step 19:
n1 < # x0 ∧ n0 = n1 ∨ # x0 ≤ n0 ∧ # x0 ≤ n1 ∧ x1 = [] ∨ # x0 ≤ n0 ∧ n1 = # x0 ∧ x1 ≠ [] ⊦ x0 = (n1 ≤ # x0 + # x1 ⊃ firstn(n1, x0 + x1);x0 + x1), n0 ≤ # x0

Rule argument

pos: right 1

Proof step 11 from proof prefixn-eq-prefixn-app (previous step)

Heuristic if-then-else split applied rule case distinction

And obtained the following new goals

Rule argument

Additional information